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\title{Chapter 18: Preservation of Holonomy}
\author{SCC ET AL}
%\institute[XX大学]{XX大学\quad 数学与统计学院\quad 数学与应用数学专业}
%\date{2025年6月}

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% Section 0
%\section{INTRO.}
\begin{frame}[allowframebreaks]{intro. }
    

We have seen in the previous chapters that {\color{red}{\color{red}holonomic}} modules are preserved by inverse images under projections and by direct images under embeddings. 

However, as we also saw, inverse images under embeddings and direct images under projections do not preserve the fact that a module is finitely generated. 

Fortunately, though, {\color{red}{\color{red}holonomic}} modules are preserved by all kinds of inverse and direct images. 

The proof of this result will use all the machinery that we have developed so far. 

It gives yet one more way to construct examples of {\color{red}{\color{red}holonomic}} modules. 

We retain the notations of 14.1.2.

\end{frame}

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% Section 1
\section{Inverse Images}
\begin{frame}[allowframebreaks]{A. }

The key to the results in this chapter is a decomposition of polynomial maps in terms of embeddings and projections. 

The idea goes back to A. Grothendieck.

Let $F: X \to Y$ be a polynomial map. 

We may decompose $F$ as a composition of three polynomial maps: a projection, an embedding and an isomorphism. 

The maps are the following. 

The projection is $\pi: X \times Y \to Y$, defined by $\pi(X,Y) = Y$. 

The isomorphism is $G: X \times Y \to X \times Y$ where $G(X,Y) = (X,Y + F(X))$. 

Finally, the embedding is $\iota: X \to X \times Y$, defined by $\iota(X) = (X,0)$. 

One can immediately check that $F = \pi \cdot G \cdot \iota$.


Let us consider the effect of an inverse image under $F$ on the {\color{red}{\color{red}holonomic}} $A_m$-module $M$. 

By Theorem 15.2.1,

\begin{equation}
F{\,}^* M \cong \iota^*(G{\,}^*(\pi^* M)).
\end{equation}

It follows from Corollary 14.3.2 and Corollary 15.3.3 that $G{\,}^*(\pi^* M)$ is {\color{red}{\color{red}holonomic}} if $M$ is {\color{red}{\color{red}holonomic}}. 

Hence $F{\,}^* M$ will be a {\color{red}{\color{red}holonomic}} module if we can prove that $\iota^*$ preserves {\color{red}holonomy}. 

We shall now do this.

We will begin with the standard embedding $\iota: X \to X \times K$. 

The coordinates of $X \times K$ will be denoted by $x_1, \ldots, x_n, y$ and $H$ will stand for the hyperplane $y = 0$.


%\subsection{Lemma}

\textbf{1.1 Lemma.} Let $M$ be a left $A_{n+1}$-module. 

Put $M{\,}' = M/\Gamma_H M$. 

Then
\begin{equation}
\iota^* M \cong \iota^* M{\,}'.
\end{equation}

\textbf{Proof:} The inverse image $\iota^* M$ is isomorphic to $M/yM$. 

Similarly, $\iota^* M{\,}' \cong M{\,}'/yM{\,}'$. 

Since $M{\,}'$ is a quotient of $M$, there exists a surjective map of $A_n$-modules,
\begin{equation}
\phi : M/yM \to M{\,}'/yM{\,}'.
\end{equation}

Let $u \in M$ and suppose that $\phi(u + yM) = 0$. 

Then $u \in yM + \Gamma_H M$. 

Since $y\Gamma_H M = \Gamma_H M$ by Corollary 17.2.5; we conclude that $u \in yM$ and that $\phi$ is an isomorphism.

%\subsection{Lemma}

\textbf{1.2 Lemma.} Let $M$ be a {\color{red}{\color{red}holonomic}} $A_{n+1}$-module {\color{blue}which does not contain any non-zero element with support on $H$}. 

Then $\iota^* M$ is a {\color{red}{\color{red}holonomic}} $A_n$-module whose multiplicity cannot exceed the multiplicity of $M$.

\textbf{Proof:} Let $\Gamma$ be a good filtration for $M$. 

Put
\begin{equation}
\Omega_j = (\Gamma_j + yM)/yM.
\end{equation}

Note that we are using a filtration of $M$ - an $A_{n+1}$-module - to construct a filtration of $M/yM$ - an $A_n$-module. 

It is clear that $\Omega_j \subseteq \Omega_{j+1}$ and that $\bigcup_{j \geq 0} \Omega_j = M/yM$. 

Since $B_i(A_n) \subseteq B_i(A_{n+1})$, we have that $B_i(A_n)\Gamma_j \subseteq \Gamma_{i+j}$. 

Thus
\begin{equation}
B_i(A_n)\Omega_j \subseteq \Omega_{i+j}.
\end{equation}

Hence $\Omega = \{\Omega_j : j \geq 0\}$ is a filtration of the $A_n$-module $M/yM$. 

Note that we do not know whether this filtration is good.

By the third homomorphism theorem for vector spaces,
\begin{equation}
\Omega_j \cong \Gamma_j / (\Gamma_j \cap yM).
\end{equation}

Therefore,
\begin{equation}
\dim_K \Omega_j = \dim_K \Gamma_j - \dim_K (\Gamma_j \cap yM).
\end{equation}

But $y\Gamma_{j-1} \subseteq \Gamma_j \cap yM$, and so
\begin{equation}
\dim_K \Omega_j \leq \dim_K \Gamma_j - \dim_K (y\Gamma_{j-1}).
\end{equation}

Since $M$ does not have any element supported on $H$, the map $\Gamma_{j-1} \to y\Gamma_{j-1}$ given by right multiplication by $y$ is injective. 

Hence $\dim_K(y\Gamma_{j-1}) = \dim_K(\Gamma_{j-1})$ and so
\begin{equation}
\dim_K \Omega_j \leq \dim_K \Gamma_j - \dim_K (\Gamma_{j-1}).
\tag{1.3}
\end{equation}

Now let $\chi(t,M,\Gamma)$ be the Hilbert polynomial of $M$ with respect to the filtration $\Gamma$. 

For $j \gg 0$, it follows from (1.3) that
\begin{equation}
\dim_K \Omega_j \leq \chi(j,M,\Gamma) - \chi(j-1,M,\Gamma).
\end{equation}

The polynomials $\chi(t,M,\Gamma)$ and $\chi(t-1,M,\Gamma)$ have the same leading term,
\begin{equation}
m(M)t^n/n!,
\end{equation}
where $m(M)$ stands for the multiplicity of $M$. 

Hence their difference has leading term $m(M)t^{n-1}/(n-1)!$. 

Thus there exists $c \in \mathbb{Q}$ such that
\begin{equation}
\dim_K \Omega_j \leq \frac{m(M)j^{n-1}}{(n-1)!} + c(j+1)^{n-2}.
\end{equation}

We conclude from Lemma 10.3.1 that $M/yM$ is a {\color{red}{\color{red}holonomic}} module whose multiplicity cannot exceed $m(M)$.

%\subsection{Theorem}

\textbf{1.4 Theorem.} Let $\iota: X \to X \times Y$ be the standard embedding and let $M$ be a {\color{red}{\color{red}holonomic}} $A_{m+n}$-module. 

Then $\iota^* M$ is a {\color{red}{\color{red}holonomic}} $A_n$-module.

\textbf{Proof:} Recall that $Y = K^m$. 

We proceed by induction on $m$. 

Suppose first that $m=1$. 

Denote by $H$ the hyperplane $y_1 = 0$, and put $M{\,}' = M/\Gamma_H M$. 

The module $M{\,}'$ is a quotient of a {\color{red}{\color{red}holonomic}} module, hence is {\color{red}{\color{red}holonomic}}. 

Since $M{\,}'$ does not contain any element supported on $H$, it follows from Lemma 1.2 that $\iota^* M{\,}'$ is {\color{red}holonomic}. 

But by Lemma 1.1, $\iota^* M \cong \iota^* M{\,}'$, hence $\iota^* M$ is also {\color{red}holonomic}.


Now the standard embedding $\iota: X \to X \times Y$ may be written as a composition of two standard embeddings, namely $\iota_1: X \to X \times K^{n-1}$ and $\iota_2: X \times K^{n-1} \to X \times Y$. 

By Theorem 15.2.1,
\begin{equation}
\iota^* M = \iota_1^* \iota_2^* M.
\end{equation}

The result follows by induction.

Let us now put all these results together.

%\subsection{Theorem}

\textbf{1.5 Theorem.} Let $F: X^n \to Y^m$ be a polynomial map. 

If $M$ is a {\color{red}holonomic} $A_m$-module, then $F{\,}^* M$ is a {\color{red}holonomic} $A_n$-module.


\end{frame}

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% Section 2
\section{Direct Images}
\begin{frame}[allowframebreaks]{B. }

That {\color{red}holonomy} is preserved under direct images is proved in a very similar way to the inverse image case. 

However we must first tidy up some loose ends.

%\subsection{Theorem}

\textbf{2.1 Theorem.} Let $F: X \to Y$ and $G: Y \to Z$ be two polynomial maps. 

Let $M$ be a left $A_n$-module. 

Then
\begin{equation}
(GF)_* M \cong G_* F_* M.
\end{equation}

\textbf{Proof:} It follows from Theorem 15.2.1 that $(GF)^*(A_r) \cong F{\,}^* G{\,}^*(A_r)$. 

In the notation of Ch. 16, §1, this is equivalent to

\begin{equation}
D_{X \rightarrow Z} \cong D_{X \rightarrow Y} \otimes_{A_m} D_{Y \rightarrow Z}.
\end{equation}

Applying Lemma 16.2.2 to this isomorphism we get that

\begin{equation}
D_{Z \leftarrow Y} \otimes_{A_m} D_{Y \leftarrow X} \cong D_{Z \leftarrow X}.
\end{equation}

The theorem is an immediate consequence of this formula.

Now, in the notation of §1, we have that the polynomial map $F: X \to Y$ can be written as $F = \pi \cdot G \cdot \iota$. 

Let $M$ be a {\color{red}holonomic} $A_m$-module. 

By Theorem 2.1,
\begin{equation}
F_* M = \pi_* (G_* (\iota_* M)).
\end{equation}

We already know, from Corollaries 16.3.3 and 17.1.3, that $G_* (\iota_* M)$ is {\color{red}holonomic}. 

In order to show that $F_* M$ is {\color{red}holonomic}, it is enough to show that $\pi_*$ preserves {\color{red}holonomy}.

%\subsection{Theorem}

\textbf{2.2 Theorem.} Let $\pi: X \times Y \to Y$ be the projection $\pi(X,Y) = Y$. 

Let $M$ be a {\color{red}holonomic} $A_{m+n}$-module. 

Then $\pi_* M$ is a {\color{red}holonomic} $A_m$-module.

\textbf{Proof:} We know from Ch. 16, §3, that
\begin{equation}
\pi_* M \cong M / \sum_{1}^{n} \partial_{z_i} M.
\end{equation}

By Proposition 5.2.1 this is the Fourier transform of $M / \sum_{1}^{n} x_i M$. 

Therefore the modules $\pi_* M$ and $M / \sum_{1}^{n} x_i M$ have the same dimension by Proposition 9.2.2. 

However, $M / \sum_{1}^{n} x_i M$ is isomorphic to $\iota^* M$ where $\iota: Y \to X \times Y$ is the embedding $\iota(Y) = (0,Y)$. 

By Theorem 1.4, $\iota^* M$ is {\color{red}holonomic}; hence so is $\pi_* M$.

Summing up, we have proved the following theorem.

%\subsection{Theorem}

\textbf{2.3 Theorem.} Let $F: X \to Y$ be a polynomial map. 

If $M$ is a {\color{red}holonomic} $A_n$-module, then $F_* M$ is a {\color{red}holonomic} $A_m$-module.

\end{frame}

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% Section 3
\section{Categories and Functors}
\begin{frame}[allowframebreaks]{C. }

We now introduce direct and inverse images as functors and generalize our previous version of Kashiwara's theorem using categories. 

For that we assume that the reader is familiar with the language of category theory. 

All that is required here can be found in [Cohn 79].


We shall be dealing with three categories. 

The category of left $A_n$-modules and module homomorphisms will be denoted by $\mathcal{M}^n$, and the full subcategory of finitely generated left $A_n$-modules by $\mathcal{M}_f^n$. 

The full subcategory of $\mathcal{M}^n$ whose objects are {\color{red}holonomic} left $A_n$-modules will be denoted by $\mathcal{H}^n$. 

Since {\color{red}holonomic} modules are finitely generated, it follows that $\mathcal{H}^n$ is a full subcategory of $\mathcal{M}_f^n$. 

Note that these are all abelian categories.


Let $F: X \to Y$ be a polynomial map. 

The inverse image $F{\,}^*$ is a functor,

\begin{equation}
F{\,}^*: \mathcal{M}^m \to \mathcal{M}^n.
\end{equation}

If $M$ is an object in $\mathcal{M}^m$ then
\begin{equation}
F{\,}^* M \cong D_{X \leftarrow Y} \otimes_{A_m} M.
\end{equation}

Thus $F{\,}^*$ is a right exact functor by [Cohn 79, §4.3, Proposition 3]. 

In general, this functor is not exact. 

For example, let $\iota: K \to K^2$ be the standard embedding: $\iota(x) = (x,0)$. 

Recall that $\iota^*(A_2) = A_2/x_2 A_2$. 

Let $\phi: A_2 \to A_2$ be the map defined by $\phi(d) = dx_2$. 

Applying the functor $\iota^*$ we get a homomorphism of $A_1$-modules,
\begin{equation}
\iota^*(\phi): A_2/x_2 A_2 \to A_2/x_2 A_2.
\end{equation}

But $1 + x_2 A_2$ is mapped to zero by $\iota^*(\phi)$; hence it cannot be injective. 

Thus $\iota^*$ is not an exact functor.


An important special case occurs when the polynomial map is a projection. 

Define $\pi: X \times Y \to Y$ by $\pi(X,Y) = Y$. 

Then
\begin{equation}
D_{X \times Y \leftarrow Y} \cong A_{m+n} / \sum_{1}^{n} A_{m+n} \partial_{x_i}
\end{equation}
is free as a right $A_m$-module: the monomials in the $x_i$'s form a basis for this module. 

Since free modules are flat, we conclude that the functor
\begin{equation}
\pi^*: \mathcal{M}^m \to \mathcal{M}^{m+n}
\end{equation}
is exact. 

We have also seen that $\pi^*$ preserves noetherianness. 

Hence we may restrict it to a functor,
\begin{equation}
\pi^*: \mathcal{M}_f^m \to \mathcal{M}_f^{m+n}.
\end{equation}

This last statement does not hold true for general polynomial maps: it is false for embeddings, for example; see Ch. 15, §1. 

On the other hand, if $F: X \to Y$ is any polynomial map, then by Theorem 1.5 the inverse image $F{\,}^*$ restricts to a functor from $\mathcal{H}^m$ to $\mathcal{H}^n$. 

So the behaviour of the inverse image functor in the {\color{red}holonomic} category is up to expectations.


For direct images we have a similar situation. 

Given a polynomial map $F: X \to Y$, the direct image $F_*$ determines a functor,
\begin{equation}
F_*: \mathcal{M}^n \to \mathcal{M}^m.
\end{equation}

Since $F_*$ is also defined by a tensor product, the direct image is right exact. 

It is not in general exact. 

Although the direct image does not always preserve finitely generated modules, it preserves {\color{red}holonomic} modules by Theorem 2.3.

Let us turn to direct images under embeddings. 

Denoting the coordinates of $X \times K$ by $x_1, \ldots, x_n, y$, let $\iota: X \to X \times K$ be the standard embedding $\iota(X) = (X,0)$. 

Then
\begin{equation}
D_{X \times K \leftarrow X} \cong A_{n+1} / A_{n+1} y
\end{equation}
is free as a right $A_n$-module: the monomials in $\partial_y$ form a basis. 

Hence the functor
\begin{equation}
\iota_*: \mathcal{M}^n \to \mathcal{M}^{n+1}
\end{equation}
is exact. 

Since $D_{X \times K \leftarrow X}$ is finitely generated as a left $A_{n+1}$-module, the functor $\iota_*$ restricts to a functor $\mathcal{M}_f^n \to \mathcal{M}_f^{n+1}$.

Now let $H$ be the hyperplane defined by the equation $y = 0$. 

Let $\mathcal{M}^{n+1}(H)$ be the full subcategory of $\mathcal{M}^{n+1}$ of modules with support on $H$. 

Thus, $M$ is an object in $\mathcal{M}^{n+1}(H)$ if and only if $\Gamma_H M = M$. 

In Lemma 17.1.4 we showed that if $M$ is an object in $\mathcal{M}^n$, then $\iota_* M$ has support on $H$. 

This leads to a categorical version of Kashiwara's theorem.

%\subsection{Theorem}

\textbf{3.1 Theorem.} Let $\iota: X \to X \times K$ be the embedding $\iota(X) = (X,0)$ and $H$ the hyperplane of equation $y = 0$. 

The direct image functor,
\begin{equation}
\iota_* : \mathcal{M}^n \to \mathcal{M}^{n+1}(H),
\end{equation}
is an equivalence of categories.

\textbf{Proof:} Consider the functor
\begin{equation}
\mathcal{K}: \mathcal{M}^{n+1} \to \mathcal{M}^n
\end{equation}
defined on a left $A_{n+1}$-module $M$ by $\mathcal{K}(M) = \ker_M H$. 

By Theorem 17.2.4 we have that
\begin{equation}
\iota_*(\mathcal{K}(M)) = \Gamma_H M.
\end{equation}

If $M$ has support on $H$, then $\iota_*(\mathcal{K}(M)) = M$.

On the other hand, let $N$ be any left $A_n$-module. 

Then
\begin{equation}
\iota_* N \cong K[\partial_y] \widehat{\otimes} N
\end{equation}
by Ch. 17, §1. 

Since the only elements of $K[\partial_y]$ that are annihilated by $y$ are the constants, we conclude that
\begin{equation}
\ker_{\iota_* N} H = K \widehat{\otimes} N \cong N
\end{equation}
as $A_n$-modules. 

Hence $\mathcal{K}(\iota_* N) \cong N$.


We leave it to the reader to check that the natural isomorphisms defined above have the expected behaviour on maps.

This may be generalized to embeddings in higher dimension. 

Let $\iota: X \to X \times Y$ be the embedding $\iota(X) = (X,0)$. 

We will identify $Y$ with the subspace of equations $y_1 = \cdots = y_m = 0$. 

Let $M$ be an object in $\mathcal{M}^{m+n}$. 

Write $H_i$ for the hyperplane $y_i = 0$. 

We say that $M$ has \textit{support} on $Y$ if it has support on $H_i$, for every $i = 1, \ldots, m$. 

The full subcategory of $A_{m+n}$-modules with support on $Y$ will be denoted by $\mathcal{M}^{m+n}(Y)$.

%\subsection{Corollary}

\textbf{3.2 Corollary.} The functor
\begin{equation}
\iota_* : \mathcal{M}^n \to \mathcal{M}^{m+n}(Y)
\end{equation}
is an equivalence of categories.

\textbf{Proof:} Recall that $Y = K^m$. 

We proceed by induction on $m$. 

The case $m=1$ has been proved in the theorem. 

Let $W = K^{m-1}$. 

There are embeddings $\iota_1: X \to X \times W$ and $\iota_2: X \times W \to X \times Y$ such that $\iota = \iota_2 \cdot \iota_1$. 

It follows by induction that $(\iota_1)_*$ and $(\iota_2)_*$ are equivalences of categories. 

Since, by Theorem 2.1,
\begin{equation}
\iota_* = (\iota_2)_* (\iota_1)_*
\end{equation}
we have that $\iota_*$ is an equivalence of categories.


Since the direct image under embeddings preserves finitely generated modules, we may replace the categories in Corollary 3.2 by the corresponding full subcategories of finitely generated modules. 

A similar statement holds for the {\color{red}holonomic} categories, see Exercise 4.5.

Kashiwara's theorem can be used to prove a structure theorem for $A_n$-modules with support on the origin.

%\subsection{Corollary}

\textbf{3.3 Corollary.} If $M$ is a finitely generated left $A_n$-module with support on the origin, then there exists an integer $r \geq 0$ such that
\begin{equation}
M \cong (K[\partial_1, \ldots, \partial_n])^r
\end{equation}

\textbf{Proof:} Let $\iota: \{0\} \to X$ be the standard embedding: $\iota(0) = 0$. 

Note first that it follows from the definitions that
\begin{equation}
K[\partial_1, \ldots, \partial_n] \cong \iota_*(K).
\end{equation}

Hence $K[\partial_1, \ldots, \partial_n]$ has support on every hyperplane $H_i$, for $1 \leq i \leq n$, by Lemma 17.1.4. 

Therefore it has support on $H_1 \cap \cdots \cap H_n = \{0\}$.

On the other hand, if $M$ has support on the origin then $\mathcal{K}(M)$ is a finitely generated module over $A_0 = K$. 

Hence $\mathcal{K}(M) \cong K^r$, for some $r \geq 0$. 

But $\iota_*(\mathcal{K}(M)) = M$, whilst
\begin{equation}
\iota_*(K^r) \cong (K[\partial_1, \ldots, \partial_n])^r.
\end{equation}


\end{frame}

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% Section 4
\section{Exercises}
\begin{frame}[allowframebreaks]{D. }

\textbf{Exercise 4.1}
Let $M, N$ be {\color{red}holonomic} $A_n$-modules. 

Show that $M \otimes_{K[X]} N$ is a {\color{red}holonomic} $A_n$-module.

\textbf{Hint:} Exercise 15.4.5.

\newpage

\textbf{Exercise 4.2}
Let $p \in K[X]$ be a non-zero polynomial and let $M$ be a {\color{red}holonomic} $A_n$-module. 

Show that $M[p^{-1}] = K[X,p^{-1}] \otimes_{K[X]} M$ is a {\color{red}holonomic} $A_n$-module.

\newpage

\textbf{Exercise 4.3}
Give an example to show that the direct image functor is not exact under projections.

\newpage

\textbf{Exercise 4.4}
Let $n \leq k \leq 2n$ be positive integers. 

Use Kashiwara's theorem and induction to construct an $A_n$-module of dimension $k$.

\newpage

\textbf{Exercise 4.5}
As in §3, let us identify $Y$ with the linear subspace of equations $y_1 = \cdots = y_m = 0$ in $X \times Y$. 

Let $\mathcal{H}^{m+n}(Y)$ be the category of {\color{red}holonomic} left $A_{m+n}$-modules with support on $Y$. 

Show that this category is equivalent to $\mathcal{H}^n$.


\end{frame}

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\end{document}


